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<h1>Dome Radius</h1>

<p>In this chapter we will develop the formula to calculate the dome radius, DR.</p>

<p class="imgbox"><a name="fig-058"></a><img src="../img/fig-058.png" style="max-width: 400px;"/><br />Figure 58: Calculating the dome radius.</p>

<h2>What We Know</h2>

<pre class="formula">
(1)	P1P2 = BL = the body length
(2)	P1P5 = BP = bridge position
(3)	P5P6 = DH = dome height at bridge position
</pre>

<h2>What We Want</h2>

<pre class="formula">
(4)	P0P1 = P0P2 = P0P3 = P0P6 = DR = dome radius
</pre>

<h2>Calculation Path</h2>

<p>We start out with the most known triangle: P1-P5-P6. </p>

<p class="imgbox"><a name="fig-059"></a><img src="../img/fig-059.png" style="max-width: 400px;"/><br />Figure 59</p>

<p>It has a right angle at P5, so we can calculate the length of P1P6:</p>

<pre class="formula">
(5)	P1P6<sup>2</sup> = P1P5<sup>2</sup> + P5P6<sup>2</sup>
(6)	P1P6 = sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )
</pre>

<p>Now let P8 divide P1-P6 into halves. Then we create a perpendicular bisector of P1-P6 through this point. This line also goes throuth the dome circle origin, P0:</p>

<p class="imgbox"><a name="fig-060"></a><img src="../img/fig-060.png" style="max-width: 400px;"/><br />Figure 60</p>

<pre class="formula">
(7)	P1P7 = P7P6 = P1P6 / 2
</pre>

<p>Currently, we cannot directly calculate the square triangles P0-P8-P1 or P0-P8-P6, because we don't know the length of the sides except one, and we don't know the angles either.</p>

<p>But we can calculate the angle a1 in the triangle P5-P1-P6, because it has a right angle at P5:</p>

<pre class="formula">
(8)	sin a1 = P5P6 /P1P6
(9)	a1 = arcsin( P5P6 /P1P6 ) 
</pre>

<p>We need another helper line, a bisector of the triangle P1-P0-P2, which goes through P4 and intersects the dome circle at the maximum dome height in P3. This line also crosses the reference line P1-P2 at a right angle.</p>

<p class="imgbox"><a name="fig-061"></a><img src="../img/fig-061.png" style="max-width: 400px;"/><br />Figure 61</p>

<p>As defined, the point P4 divides P1-P2 into half:</p>

<pre class="formula">
(10)	P1P4 = P1P2 / 2
</pre>

<p>This helper also creates the point P9 at the intersection with P1-P6, which constitutes the interesting triangle P0-P8-P9, with a right angle at P8.</p>

<p>We also can calculate the including angle a2 of P1-P9-P4 immediately:</p>

<pre class="formula">
(11)	a2 = 90 - a1
</pre>

<p>This angle allows us later to determine triangle P0-P8-P9. But we need to determine the side P8-P9 first. Because P4-P9 is parallel to P5-P6, we can set the following lines into relation, and isolate P1-P9:</p>

<pre class="formula">
(12)	P1P4 / P1P5 = P1P9 / P1P6
(13)	P1P9 = P1P6 * P1P4 / P1P5
</pre>

<p>Because we already know the length of P1-P6 from (6), and we know that P1-P8 is half of that (7), we can determine P7P9 now:</p>

<pre class="formula">
(14)	P1P9 = P1P7 + P7P9
(15)	P7P9 = P1P9 - P1P7 = P1P9 - P1P6 / 2 
</pre>

<p>Substitution of P1P9 with equation (13) gives</p>

<pre class="formula">
(16)	P7P9 = P1P6 * P1P4 / P1P5 - P1P6 / 2 
</pre>

<p>Now we can calculate P0-P8:</p>

<pre class="formula">
(17)	tan a2 = P0P7 / P7P9
(18)	P0P7 = P7P9 * tan a2
(19)	P0P7 = (P1P6 * P1P4 / P1P5 - P1P6 / 2) * tan a2
</pre>

<p>Now we can find the desired dome radius P0-P1 from the triangle P1-P8-P0:</p>

<pre class="formula">
(20)	P0P1 = sqrt( P1P7<sup>2</sup> + P0P7<sup>2</sup> )
</pre>

<p>Substitution of P0-P8 from (19) and P1P7 from (7) gives us</p>

<pre class="formula">
(21)	P0P1 = sqrt( (P1P6/2)<sup>2</sup> 
	+ ( (P1P6 * P1P4 / P1P5 - P1P6 / 2) * tan a2)<sup>2</sup> )
</pre>

<p>From equation (10) we know that P1P4 is half the body length, and with the substitution of P1P6 from (6) we get</p>

<pre class="formula">
(22)	P0P1 = sqrt( (P1P6/2)<sup>2</sup> 
	+ ( (P1P6 * P1P2 / 2 / P1P5 - P1P6 / 2) * tan a2)<sup>2</sup> )
(23)	P0P1 = sqrt((( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) / 2 )<sup>2</sup> 
	+ ((( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) * P1P2 / 2 / P1P5 
	- ( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) / 2) * tan a2)<sup>2</sup> )
</pre>

<p>The last step substitutes tan a2 from (11), (9) and (5):</p>

<pre class="formula">
(24)	tan a2 = tan ( 90 - a1 ) = tan( 90 - arcsin( P5P6 /P1P6 ))
(25)	tan a2 = tan( 90 - arcsin( P5P6 /( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> ))))
</pre>

<p>Introduced into (23) makes a real bandworm:</p>

<pre class="formula">
(26)	P0P1 = sqrt((( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) / 2 )<sup>2</sup> 
	+ ((( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) * P1P2 / 2 / P1P5 
	- ( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )) / 2) 
	* tan( 90 - arcsin( P5P6 /( sqrt( P1P5<sup>2</sup> + P5P6<sup>2</sup> )))))<sup>2</sup> )
</pre>

<p>Introducing the original names from equations (1) to (4) gives us the final formula:</p>

<pre class="formula final">
(27)	DR = sqrt((( sqrt( BP<sup>2</sup> + DH<sup>2</sup> )) / 2 )<sup>2</sup> 
	+ ((( sqrt( BP<sup>2</sup> + DH<sup>2</sup> )) * BL / 2 / BP 
	- ( sqrt( BP<sup>2</sup> + DH<sup>2</sup> )) / 2) 
	* tan( 90 - arcsin( DH /( sqrt( BP<sup>2</sup> + DH<sup>2</sup> )))))<sup>2</sup> )
</pre>

<p>Admittedly, this formula is rather counter-intuitive, but a testing against the other equations proves the correctness.</p>

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